Class 7: Maths Chapter 12 solutions. Complete Class 7 Maths Chapter 12 Notes.
Contents
Selina Class 7 ICSE Solutions Mathematics : Chapter 12- Simple Linear Equations (Including Word Problems)
Selina 7th Maths Chapter 12, Class 7 Maths Chapter 12 solutions
Exercise 12A page: 139
Solve the following questions:
1. x + 5 = 10
Solution:
x + 5 = 10
So we get
x = 10 – 5 = 5
2. 2 + y = 7
Solution:
2 + y = 7
So we get
y = 7 – 2 = 5
3. a – 2 = 6
Solution:
a – 2 = 6
So we get
a = 6 + 2 = 8
4. x – 5 = 8
Solution:
x – 5 = 8
So we get
x = 8 + 5 = 13
5. 5 – d = 12
Solution:
5 – d = 12
So we get
– d = 12 – 5 = 7
d = -7
6. 3p = 12
Solution:
3p = 12
So we get
p = 12/3 = 4
7. 14 = 7m
Solution:
14 = 7m
So we get
m = 14/7 = 2
8. 2x = 0
Solution:
2x = 0
So we get
x = 0/2 = 0
9. x/9 = 2
Solution:
x/9 = 2
So we get
x = 2 × 9 = 18
10. y/-12 = -4
Solution:
y/-12 = -4
So we get
y = -4 × -12 = 48
11. 8x – 2 = 38
Solution:
8x – 2 = 38
So we get
8x = 38 + 2
8x = 40
x = 40/8 = 5
12. 2x + 5 = 5
Solution:
2x + 5 = 5
So we get
2x = 5 – 5
2x = 0
x = 0/2 = 0
13. 5x – 1 = 74
Solution:
5x – 1 = 74
So we get
5x = 74 + 1
5x = 75
x = 75/5 = 15
14. 14 = 27 – x
Solution:
14 = 27 – x
So we get
x = 27 – 14 = 13
15. 10 + 6a = 40
Solution:
10 + 6a = 40
So we get
6a = 40 – 10
6a = 30
a = 30/6 = 5
Exercise 12B page: 140
Solve:
1. 8y – 4y = 20
Solution:
8y – 4y = 20
By further calculation
4y = 20
So we get
y = 20/4 = 5
2. 9b – 4b + 3b = 16
Solution:
9b – 4b + 3b = 16
By further calculation
8b = 16
So we get
b = 16/8 = 2
3. 5y + 8 = 8y – 18
Solution:
5y + 8 = 8y – 18
By further calculation
8y – 5y = 8 + 18
So we get
3y = 26
y = 26/3 = 8 2/3
4. 6 = 7 + 2p – 5
Solution:
6 = 7 + 2p – 5
By further calculation
2p = 6 – 7 + 5
So we get
2p = 4
p = 4/2 = 2
5. 8 – 7x = 13x + 8
Solution:
8 – 7x = 13x + 8
By further calculation
13x + 7x = 8 – 8
20x = 0
x = 0/20 = 0
6. 4x – 5x + 2x = 28 + 3x
Solution:
4x – 5x + 2x = 28 + 3x
By further calculation
4x – 5x + 2x – 3x = 28
So we get
– 2x = 28
x = 28/-2 = – 14
7. 9 + m = 6m + 8 – m
Solution:
9 + m = 6m + 8 – m
By further calculation
6m – m – m = 9 – 8
4m = 1
So we get
m = 1/4
8. 24 = y + 2y + 3 + 4y
Solution:
24 = y + 2y + 3 + 4y
By further calculation
24 – 3 = 7y
21 = 7y
So we get
y = 21/7 = 3
9. 19x + 13 – 12x + 3 = 23
Solution:
19x + 13 – 12x + 3 = 23
By further calculation
7x = 23 – 16
So we get
7x = 7
x = 7/7 = 1
10. 6b + 40 = – 100 – b
Solution:
6b + 40 = – 100 – b
By further calculation
7b = – 100 – 40
7b = -140
So we get
b = -140/7 = – 20
11. 6 – 5m – 1 + 3m = 0
Solution:
6 – 5m – 1 + 3m = 0
By further calculation
5 – 2m = 0
So we get
2m = 5
m = 5/2 = 2 1/2
12. 0.4x – 1.2 = 0.3x + 0.6
Solution:
0.4x – 1.2 = 0.3x + 0.6
By further calculation
0.1x = 1.8
Multiply and divide both numerator and denominator by 10
1/10x = 18/10
By cross multiplication
x = 18/10 × 10/1 = 18
13. 6 (x + 4) = 36
Solution:
6 (x + 4) = 36
By further calculation
6x + 24 = 36
So we get
6x = 36 – 24
6x = 12
x = 12/6 = 2
14. 9 (a + 5) + 2 = 11
Solution:
9 (a + 5) + 2 = 11
By further calculation
9a + 45 + 2 = 11
So we get
9a = 11 – 47
9a = – 36
a = -36/9 = – 4
15. 4 (x – 2) = 12
Solution:
4 (x – 2) = 12
By further calculation
4x – 8 = 12
So we get
4x = 20
x = 20/4 = 5
Exercise 12C page: 141
Solve:
1. x/2 + x = 9
Solution:
It is given that
x/2 + x/1 = 9
Taking LCM
(x + 2x)/ 2 = 9
By cross multiplication
x + 2x = 9 × 2
3x = 18
So we get
x = 18/3 = 6
2. x/5 + 2x = 33
Solution:
It is given that
x/5 + 2x/1 = 33
Taking LCM
(x + 10x)/5 = 33
11x/5 = 33
By cross multiplication
11x = 5 × 33 = 165
So we get
x = 165/11 = 15
3. 3x/4 + 4x = 38
Solution:
It is given that
3x/4 + 4x/1 = 38
Taking LCM
(3x + 16x)/ 4 = 38
19x/4 = 38
By cross multiplication
19x = 38 × 4 = 152
So we get
x = 152/19 = 8
4. x/2 + x/5 = 14
Solution:
It is given that
x/2 + x/5 = 14
Taking LCM
(5x + 2x)/ 10 = 14
7x/10 = 14
By cross multiplication
7x = 14 × 10 = 140
So we get
x = 140/7 = 20
5. x/3 – x/4 = 2
Solution:
It is given that
x/3 – x/4 = 2
Taking LCM
(4x – 3x)/ 12 = 2
x/12 = 2
By cross multiplication
x = 12 × 2 = 24
6. y + y/2 = 7/4 – y/4
Solution:
It is given that
y/1 + y/2 = 7/4 – y/4
y/1 + y/2 + y/4 = 7/4
Taking LCM
(4y + 2y + y)/4 = 7/4
7y/4 = 7/4
7y = 7
So we get
y = 7/7 = 1
7. 4x/3 – 7x/3 = 1
Solution:
It is given that
4x/3 – 7x/3 = 1
By further calculation
-3x/3 = 1
So we get
-x = 1
x = – 1
8. (1/2)m + (3/4)m – m = 2.5
Solution:
It is given that
(1/2)m + (3/4)m – m/1 = 2.5
Taking LCM
(2m + 3m – 4m)/ 4 = 2.5
m/4 = 2.5
By cross multiplication
m = 2.5 × 4 = 10
9. 2x/3 + x/2 – 3x/4 = 1
Solution:
It is given that
2x/3 + x/2 – 3x/4 = 1
Taking LCM
(8x + 6x – 9x)/ 12 = 1
5x/12 = 1
By cross multiplication
5x = 1 × 12 = 12
So we get
x = 12/5 = 2 2/5
10. 3a/4 + a/6 = 66
Solution:
It is given that
3a/4 + a/6 = 66
Taking LCM
(9a + 2a)/ 12 = 66
11a/12 = 66
By cross multiplication
11a = 66 × 12 = 792
So we get
a = 792/11 = 72
11. 2p/3 – p/5 = 35
Solution:
It is given that
2p/3 – p/5 = 35
Taking LCM
(10p – 3p)/ 15 = 35
7p/ 15 = 35
By cross multiplication
7p = 35 × 15 = 525
So we get
p = 525/7 = 75
12. 0.6a + 0.2a = 0.4a + 8
Solution:
It is given that
0.6a + 0.2a = 0.4a + 8
Multiplying and dividing both numerator and denominator by 10
6/10a + 2/10a = 4a/10 + 80/10
Taking LCM
(6a + 2a)/ 10 = (4a + 80)/ 10
6a + 2a = 4a + 80
So we get
4a = 80
a = 80/4 = 20
13. p + 1.4p = 48
Solution:
It is given that
p + 1.4p = 48
Multiplying and dividing both numerator and denominator by 10
p + 14/10p = 48
Taking LCM
(10p + 14p)/ 10 = 48
24p/10 = 48
By cross multiplication
24p = 48 × 10 = 480
So we get
p = 480/24 = 20
14. 10% of x = 20
Solution:
It is given that
10% of x = 20
We can write it as
10/100 × x = 20
x/10 = 20
By cross multiplication
x = 20 × 10 = 200
15. y + 20% of y = 18
Solution:
It is given that
y + 20% of y = 18
We can write it as
y + 20/100 × y = 18
Taking LCM
(100y + 20y)/ 100 = 18
By cross multiplication
120y = 18 × 100 = 1800
So we get
y = 1800/120 = 15
Exercise 12D page: 145
1. One-fifth of a number is 5, find the number.
Solution:
Consider the number = x
Based on the condition
(1/5)x = 5
By cross multiplication
x = 5 × 5 = 25
Hence, the number is 25.
2. Six times a number is 72, find the number.
Solution:
Consider the number = x
Based on the condition
6x = 72
So we get
x = 72/6 = 12
Hence, the number is 12.
3. If 15 is added to a number, the result is 69, find the number.
Solution:
Consider the number = x
Based on the condition
x + 15 = 69
So we get
x = 69 – 15 = 54
Hence, the number is 54.
4. The sum of twice a number and 4 is 80, find the number.
Solution:
Consider the number = x
Based on the condition
2x + 4 = 80
So we get
2x = 80 – 4 = 76
x = 76/2 = 38
Hence, the number is 38.
5. The difference between a number and one-fourth of itself is 24, find the number.
Solution:
Consider the number = x
Based on the condition
x – (1/4)x = 24
Taking LCM
(4x – x)/ 4 = 24
3x/4 = 24
By cross multiplication
x = 24 × 4/3
So we get
x = 8 × 4 = 32
Hence, the number is 32.
6. Find a number whose one-third part exceeds its one-fifth part by 20.
Solution:
Consider the number = x
Based on the condition
(1/3)x – (1/5)x = 20
Here the LCM of 3 and 5 is 15
(5x – 3x)/ 15 = 20
2x/15 = 20
So we get
x = 20 × 15/2 = 150
Hence, the number is 150.
7. A number is as much greater than 35 as is less than 53. Find the number.
Solution:
Consider the number = x
Based on the condition
x – 35 = 53 – x
By further calculation
2x = 88
So we get
x = 88/2 = 44
Hence, the number is 44.
8. The sum of two numbers is 18. If one is twice the other, find the numbers.
Solution:
Consider the first number = x
Second number = y
Based on the condition
x + y = 18 ….. (1)
x = 2y …… (2)
Now substituting the equation (2) in (1)
2y + y = 18
3y = 18
So we get
y = 18/3 = 6
Substituting the value of y in equation (2)
x = 2 × 6 = 12
Hence, the two numbers are 12 and 6.
9. A number is 15 more than the other. The sum of the two numbers is 195. Find the numbers.
Solution:
Consider the first number = x
Second number = y
Based on the condition
x = y + 15 ….. (1)
x + y = 195 …. (2)
Now substituting equation (1) in (2) we get
y + 15 + 7 = 195
2y = 195 – 15 = 180
So we get
y = 180/2 = 90
Substituting the value of y in equation (1)
x = 90 + 15 = 105
Hence, the two numbers are 105 and 90.
10. The sum of three consecutive even numbers is 54. Find the numbers.
Solution:
Consider the first even number = x
Second even number = x + 2
Third even number = x + 4
Based on the condition
x + x + 2 + x + 4 = 54
By further calculation
3x + 6 = 54
3x = 54 – 6 = 48
So we get
x = 48/3 = 16
First even number = 16
Second even number = 16 + 2 = 18
Third even number = 16 + 4 = 20
11. The sum of three consecutive odd numbers is 63. Find the numbers.
Solution:
Consider the first odd number = x
Second odd number = x + 2
Third odd number = x + 4
Based on the condition
x + x + 2 + x + 4 = 63
By further calculation
3x + 6 = 63
3x = 63 – 6 = 57
So we get
x = 57/3 = 19
First odd number = 19
Second odd number = 19 + 2 = 21
Third odd number = 19 + 4 = 23
12. A man has ₹ x from which he spends ₹ 6. If twice of the money left with him is ₹ 86, find x.
Solution:
Consider ₹ x as the total amount
Based on the condition
2x = 86
By further calculation
x = 86/2 = 43
Amount spent by him = ₹ 6
So the total money he have = 43 + 6 = ₹ 49
13. A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.
Solution:
Consider the present age of son = x years
So the present age of father = 4x years
After 20 years
Age of son = (x + 20) years
Age of father = (4x + 20) years
Based on the condition
4x + 20 = 2 (x + 20)
By further calculation
4x + 20 = 2x + 40
2x = 20
So we get
x = 10
So the present age of son = 10 years
Present age of father = 4 × 10 = 40 years
14. If 5 is subtracted from three times a number, the result is 16. Find the number.
Solution:
Consider x as the number
Based on the condition
3x – 5 = 16
By further calculation
3x = 16 + 5 = 21
So we get
x = 21/3 = 7
Hence, the number is 7.
15. Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.
Solution:
Consider the first consecutive number = x
Second consecutive number = x + 1
Third consecutive number = x + 2
Based on the condition
x + x + 1 = 15 + x + 2
By further calculation
2x + 1 = 17 + x
2x – x = 17 – 1
So we get
x = 16
First consecutive number = 16
Second consecutive number = 16 + 1 = 17
Third consecutive number = 16 + 2 = 18
Exercise 12B page: 140
Solve:
1. 8y – 4y = 20
Solution:
8y – 4y = 20
By further calculation
4y = 20
So we get
y = 20/4 = 5
2. 9b – 4b + 3b = 16
Solution:
9b – 4b + 3b = 16
By further calculation
8b = 16
So we get
b = 16/8 = 2
3. 5y + 8 = 8y – 18
Solution:
5y + 8 = 8y – 18
By further calculation
8y – 5y = 8 + 18
So we get
3y = 26
y = 26/3 = 8 2/3
4. 6 = 7 + 2p – 5
Solution:
6 = 7 + 2p – 5
By further calculation
2p = 6 – 7 + 5
So we get
2p = 4
p = 4/2 = 2
5. 8 – 7x = 13x + 8
Solution:
8 – 7x = 13x + 8
By further calculation
13x + 7x = 8 – 8
20x = 0
x = 0/20 = 0
6. 4x – 5x + 2x = 28 + 3x
Solution:
4x – 5x + 2x = 28 + 3x
By further calculation
4x – 5x + 2x – 3x = 28
So we get
– 2x = 28
x = 28/-2 = – 14
7. 9 + m = 6m + 8 – m
Solution:
9 + m = 6m + 8 – m
By further calculation
6m – m – m = 9 – 8
4m = 1
So we get
m = 1/4
8. 24 = y + 2y + 3 + 4y
Solution:
24 = y + 2y + 3 + 4y
By further calculation
24 – 3 = 7y
21 = 7y
So we get
y = 21/7 = 3
9. 19x + 13 – 12x + 3 = 23
Solution:
19x + 13 – 12x + 3 = 23
By further calculation
7x = 23 – 16
So we get
7x = 7
x = 7/7 = 1
10. 6b + 40 = – 100 – b
Solution:
6b + 40 = – 100 – b
By further calculation
7b = – 100 – 40
7b = -140
So we get
b = -140/7 = – 20
11. 6 – 5m – 1 + 3m = 0
Solution:
6 – 5m – 1 + 3m = 0
By further calculation
5 – 2m = 0
So we get
2m = 5
m = 5/2 = 2 1/2
12. 0.4x – 1.2 = 0.3x + 0.6
Solution:
0.4x – 1.2 = 0.3x + 0.6
By further calculation
0.1x = 1.8
Multiply and divide both numerator and denominator by 10
1/10x = 18/10
By cross multiplication
x = 18/10 × 10/1 = 18
13. 6 (x + 4) = 36
Solution:
6 (x + 4) = 36
By further calculation
6x + 24 = 36
So we get
6x = 36 – 24
6x = 12
x = 12/6 = 2
14. 9 (a + 5) + 2 = 11
Solution:
9 (a + 5) + 2 = 11
By further calculation
9a + 45 + 2 = 11
So we get
9a = 11 – 47
9a = – 36
a = -36/9 = – 4
15. 4 (x – 2) = 12
Solution:
4 (x – 2) = 12
By further calculation
4x – 8 = 12
So we get
4x = 20
x = 20/4 = 5
Exercise 12C page: 141
Solve:
1. x/2 + x = 9
Solution:
It is given that
x/2 + x/1 = 9
Taking LCM
(x + 2x)/ 2 = 9
By cross multiplication
x + 2x = 9 × 2
3x = 18
So we get
x = 18/3 = 6
2. x/5 + 2x = 33
Solution:
It is given that
x/5 + 2x/1 = 33
Taking LCM
(x + 10x)/5 = 33
11x/5 = 33
By cross multiplication
11x = 5 × 33 = 165
So we get
x = 165/11 = 15
3. 3x/4 + 4x = 38
Solution:
It is given that
3x/4 + 4x/1 = 38
Taking LCM
(3x + 16x)/ 4 = 38
19x/4 = 38
By cross multiplication
19x = 38 × 4 = 152
So we get
x = 152/19 = 8
4. x/2 + x/5 = 14
Solution:
It is given that
x/2 + x/5 = 14
Taking LCM
(5x + 2x)/ 10 = 14
7x/10 = 14
By cross multiplication
7x = 14 × 10 = 140
So we get
x = 140/7 = 20
5. x/3 – x/4 = 2
Solution:
It is given that
x/3 – x/4 = 2
Taking LCM
(4x – 3x)/ 12 = 2
x/12 = 2
By cross multiplication
x = 12 × 2 = 24
6. y + y/2 = 7/4 – y/4
Solution:
It is given that
y/1 + y/2 = 7/4 – y/4
y/1 + y/2 + y/4 = 7/4
Taking LCM
(4y + 2y + y)/4 = 7/4
7y/4 = 7/4
7y = 7
So we get
y = 7/7 = 1
7. 4x/3 – 7x/3 = 1
Solution:
It is given that
4x/3 – 7x/3 = 1
By further calculation
-3x/3 = 1
So we get
-x = 1
x = – 1
8. m/2 + 3m/4 – m = 2.5
Solution:
It is given that
m/2 + 3m/4 – m/1 = 2.5
Taking LCM
(2m + 3m – 4m)/ 4 = 2.5
m/4 = 2.5
By cross multiplication
m = 2.5 × 4 = 10
9. 2x/3 + x/2 – 3x/4 = 1
Solution:
It is given that
2x/3 + x/2 – 3x/4 = 1
Taking LCM
(8x + 6x – 9x)/ 12 = 1
5x/12 = 1
By cross multiplication
5x = 1 × 12 = 12
So we get
x = 12/5 = 2 2/5
10. 3a/4 + a/6 = 66
Solution:
It is given that
3a/4 + a/6 = 66
Taking LCM
(9a + 2a)/ 12 = 66
11a/12 = 66
By cross multiplication
11a = 66 × 12 = 792
So we get
a = 792/11 = 72
11. 2p/3 – p/5 = 35
Solution:
It is given that
2p/3 – p/5 = 35
Taking LCM
(10p – 3p)/ 15 = 35
7p/ 15 = 35
By cross multiplication
7p = 35 × 15 = 525
So we get
p = 525/7 = 75
12. 0.6a + 0.2a = 0.4a + 8
Solution:
It is given that
0.6a + 0.2a = 0.4a + 8
Multiplying and dividing both numerator and denominator by 10
6/10a + 2/10a = 4a/10 + 80/10
Taking LCM
(6a + 2a)/ 10 = (4a + 80)/ 10
6a + 2a = 4a + 80
So we get
4a = 80
a = 80/4 = 20
13. p + 1.4p = 48
Solution:
It is given that
p + 1.4p = 48
Multiplying and dividing both numerator and denominator by 10
p + 14/10p = 48
Taking LCM
(10p + 14p)/ 10 = 48
24p/10 = 48
By cross multiplication
24p = 48 × 10 = 480
So we get
p = 480/24 = 20
14. 10% of x = 20
Solution:
It is given that
10% of x = 20
We can write it as
10/100 × x = 20
x/10 = 20
By cross multiplication
x = 20 × 10 = 200
15. y + 20% of y = 18
Solution:
It is given that
y + 20% of y = 18
We can write it as
y + 20/100 × y = 18
Taking LCM
(100y + 20y)/ 100 = 18
By cross multiplication
120y = 18 × 100 = 1800
So we get
y = 1800/120 = 15
Exercise 12D page: 145
1. One-fifth of a number is 5, find the number.
Solution:
Consider the number = x
Based on the condition
(1/5)x = 5
By cross multiplication
x = 5 × 5 = 25
Hence, the number is 25.
2. Six times a number is 72, find the number.
Solution:
Consider the number = x
Based on the condition
6x = 72
So we get
x = 72/6 = 12
Hence, the number is 12.
3. If 15 is added to a number, the result is 69, find the number.
Solution:
Consider the number = x
Based on the condition
x + 15 = 69
So we get
x = 69 – 15 = 54
Hence, the number is 54.
4. The sum of twice a number and 4 is 80, find the number.
Solution:
Consider the number = x
Based on the condition
2x + 4 = 80
So we get
2x = 80 – 4 = 76
x = 76/2 = 38
Hence, the number is 38.
5. The difference between a number and one-fourth of itself is 24, find the number.
Solution:
Consider the number = x
Based on the condition
x – (1/4)x = 24
Taking LCM
(4x – x)/ 4 = 24
3x/4 = 24
By cross multiplication
x = 24 × 4/3
So we get
x = 8 × 4 = 32
Hence, the number is 32.
6. Find a number whose one-third part exceeds its one-fifth part by 20.
Solution:
Consider the number = x
Based on the condition
(1/3)x – (1/5)x = 20
Here the LCM of 3 and 5 is 15
(5x – 3x)/ 15 = 20
2x/15 = 20
So we get
x = 20 × 15/2 = 150
Hence, the number is 150.
7. A number is as much greater than 35 as is less than 53. Find the number.
Solution:
Consider the number = x
Based on the condition
x – 35 = 53 – x
By further calculation
2x = 88
So we get
x = 88/2 = 44
Hence, the number is 44.
8. The sum of two numbers is 18. If one is twice the other, find the numbers.
Solution:
Consider the first number = x
Second number = y
Based on the condition
x + y = 18 ….. (1)
x = 2y …… (2)
Now substituting the equation (2) in (1)
2y + y = 18
3y = 18
So we get
y = 18/3 = 6
Substituting the value of y in equation (2)
x = 2 × 6 = 12
Hence, the two numbers are 12 and 6.
9. A number is 15 more than the other. The sum of the two numbers is 195. Find the numbers.
Solution:
Consider the first number = x
Second number = y
Based on the condition
x = y + 15 ….. (1)
x + y = 195 …. (2)
Now substituting equation (1) in (2) we get
y + 15 + 7 = 195
2y = 195 – 15 = 180
So we get
y = 180/2 = 90
Substituting the value of y in equation (1)
x = 90 + 15 = 105
Hence, the two numbers are 105 and 90.
10. The sum of three consecutive even numbers is 54. Find the numbers.
Solution:
Consider the first even number = x
Second even number = x + 2
Third even number = x + 4
Based on the condition
x + x + 2 + x + 4 = 54
By further calculation
3x + 6 = 54
3x = 54 – 6 = 48
So we get
x = 48/3 = 16
First even number = 16
Second even number = 16 + 2 = 18
Third even number = 16 + 4 = 20
11. The sum of three consecutive odd numbers is 63. Find the numbers.
Solution:
Consider the first odd number = x
Second odd number = x + 2
Third odd number = x + 4
Based on the condition
x + x + 2 + x + 4 = 63
By further calculation
3x + 6 = 63
3x = 63 – 6 = 57
So we get
x = 57/3 = 19
First odd number = 19
Second odd number = 19 + 2 = 21
Third odd number = 19 + 4 = 23
12. A man has ₹ x from which he spends ₹ 6. If twice of the money left with him is ₹ 86, find x.
Solution:
Consider ₹ x as the total amount
Based on the condition
2x = 86
By further calculation
x = 86/2 = 43
Amount spent by him = ₹ 6
So the total money he have = 43 + 6 = ₹ 49
13. A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.
Solution:
Consider the present age of son = x years
So the present age of father = 4x years
After 20 years
Age of son = (x + 20) years
Age of father = (4x + 20) years
Based on the condition
4x + 20 = 2 (x + 20)
By further calculation
4x + 20 = 2x + 40
2x = 20
So we get
x = 10
So the present age of son = 10 years
Present age of father = 4 × 10 = 40 years
14. If 5 is subtracted from three times a number, the result is 16. Find the number.
Solution:
Consider x as the number
Based on the condition
3x – 5 = 16
By further calculation
3x = 16 + 5 = 21
So we get
x = 21/3 = 7
Hence, the number is 7.
15. Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.
Solution:
Consider the first consecutive number = x
Second consecutive number = x + 1
Third consecutive number = x + 2
Based on the condition
x + x + 1 = 15 + x + 2
By further calculation
2x + 1 = 17 + x
2x – x = 17 – 1
So we get
x = 16
First consecutive number = 16
Second consecutive number = 16 + 1 = 17
Third consecutive number = 16 + 2 = 18
Download PDF
Selina Class 7 ICSE Solutions Mathematics : Chapter 12- Simple Linear Equations (Including Word Problems)
Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :
- Chapter 1- Integers
- Chapter 2- Rational Numbers
- Chapter 3- Fraction (Including Problems)
- Chapter 4- Decimal Fractions (Decimals)
- Chapter 5- Exponents (Including Laws of Exponents)
- Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)
- Chapter 7- Unitary Method (Including Time and Work)
- Chapter 8- Percent and Percentage
- Chapter 9- Profit, Loss and Discount
- Chapter 10- Simple Interest
- Chapter 11- Fundamental Concepts (Including Fundamental Operations)
- Chapter 12- Simple Linear Equations (Including Word Problems)
- Chapter 13- Set Concepts
- Chapter 14- Lines and Angles (Including Construction of Angles)
- Chapter 15- Triangles
- Chapter 16- Pythagoras Theorem
- Chapter 17- Symmetry (Including Reflection and Rotation)
- Chapter 18- Recognition of Solids (Representing 3-D in 2-D)
- Chapter 19- Congruency: Congruent Triangles
- Chapter 20- Mensuration (Perimeter and Area of Plane Figures)
- Chapter 21- Data Handling
- Chapter 22- Probability
About Selina Publishers ICSE
Selina Publishers has been serving the students since 1976 and is one of the quality ICSE school textbooks publication houses. Mathematics and Science books for classes 6-10 form the core of our business, apart from certain English and Hindi literature as well as a few primary books. All these books are based upon the syllabus published by the Council for the I.C.S.E. Examinations, New Delhi. The textbooks are composed by a panel of subject experts and vetted by teachers practising in ICSE schools all over the country. Continuous efforts are made in complying with the standards and ensuring lucidity and clarity in content, which makes them stand tall in the industry.